find \\( \\frac{d y}{d t} \\)\n\\( y=(t \\sec t)^{11} \\)\n\\( \\frac{d y}{d t}= \\)

Explanation:

Step1: Apply the chain rule

The chain rule states that if (y = u^{n}), then (\frac{dy}{dt}=n\cdot u^{n - 1}\cdot\frac{du}{dt}). Here (y=(t\sec t)^{11}), so let (u = t\sec t) and (n = 11). Then (\frac{dy}{dt}=11(t\sec t)^{10}\cdot\frac{d}{dt}(t\sec t)).

Step2: Apply the product rule

The product rule (\frac{d}{dt}(uv)=u\frac{dv}{dt}+v\frac{du}{dt}). For (u = t) and (v=\sec t), (\frac{d}{dt}(t\sec t)=t\frac{d}{dt}(\sec t)+\sec t\frac{d}{dt}(t)). Since (\frac{d}{dt}(\sec t)=\sec t\tan t) and (\frac{d}{dt}(t) = 1), we have (\frac{d}{dt}(t\sec t)=t\sec t\tan t+\sec t).

Step3: Combine the results

Substitute (\frac{d}{dt}(t\sec t)) back into the expression from Step1: [ \begin{align*} \frac{dy}{dt}&=11(t\sec t)^{10}(t\sec t\tan t+\sec t)\ &=11\sec t(t\sec t)^{10}(t\tan t + 1) \end{align*} ]

Answer:

(11\sec t(t\sec t)^{10}(t\tan t + 1))